Mittag leffler function

67 Views Asked by At

I am soo sorry to ask like this question but really stuck on it I have to understand it.

$ {\mathrm{(}}\frac{d}{dz}{\mathrm{)}}^{m}\left[{{z}^{\mathit{\beta}\mathrm{{-}}{1}}{E}_{\mathit{\alpha}{\mathrm{,}}\mathit{\beta}\hspace{0.33em}}\mathrm{(}{z}^{\mathit{\alpha}}\mathrm{)}}\right]\mathrm{{=}}{z}^{\mathit{\beta}\mathrm{{-}}{m}\mathrm{{-}}{1}}{E}_{\mathit{\alpha}{\mathrm{,}}\mathit{\beta}\mathrm{{-}}{m}}{\mathrm{(}}{z}^{\mathit{\alpha}}{\mathrm{)}}\hspace{0.33em}{\mathrm{R}}{\mathrm{(}}\mathit{\beta}\mathrm{{-}}{m}{\mathrm{)}}\mathrm{{>}}{0}{\mathrm{,}}{m}\mathrm{{=}}{0}{\mathrm{,}}{1}{\mathrm{,....}} $

Where:

$ {E}_{\mathit{\alpha}{\mathrm{,}}\mathit{\beta}}{\mathrm{(}}{z}{\mathrm{)}}\mathrm{{=}}\mathop{\sum}\limits_{{k}\mathrm{{=}}{0}}\limits^{\mathrm{\infty}}{\frac{{z}^{k}}{\mathit{\Gamma}{\mathrm{(}}\mathit{\alpha}{k}\mathrm{{+}}\mathit{\beta}{\mathrm{)}}}} $

If there is any explanation it will be great .